Loosing solutions in a trig equation

[Appleton]

Homework Statement

find the values of x between 0 and 360
sin2x + cosx = 0

Homework Equations

The Attempt at a Solution

sin2x + cosx = 0
2sinxcosx = -cosx
2sinx = -1
sinx = -1/2
x = 210, 330
I'm guessing I lost the other solutions when I divided cosx by itself

 

[CAF123]

Homework Statement

find the values of x between 0 and 360
sin2x + cosx = 0

Homework Equations

The Attempt at a Solution

sin2x + cosx = 0
2sinxcosx = -cosx
2sinx = -1
sinx = -1/2
x = 210, 330
I'm guessing I lost the other solutions when I divided cosx by itself

Yes, you did. By dividing by cosx you assumed that cosx was never zero. In general, only cancel on both sides if you have a constant, independent of x.

 

[LCKurtz]

Homework Statement

find the values of x between 0 and 360
sin2x + cosx = 0

Homework Equations

The Attempt at a Solution

sin2x + cosx = 0
2sinxcosx = -cosx

##\cos x(2\sin x + 1) = 0## Deal with both factors.

 

[Millennial]

##\cos x(2\sin x + 1) = 0## Deal with both factors.

That is actually what he did, but he writes it differently, and he divides by [itex]\cos x[/itex] (oops.)

Basically, when you divide both sides of an equation by a function, you implicitly assume it is a nonzero function. For example, dividing both sides of an equation by [itex]a^x[/itex] when [itex]a\neq 0[/itex] is perfectly doable, because exponential functions are nonzero. However, a function like cosine has infinitely many zeroes, so you can't simply divide by cosine and forget about it.

You already solved half of the equation. But there are more solutions, and they are the zeroes of cosine in the interval given to you.

 

[Appleton]

Thanks for your responses. So the other solutions within the given range are 90 and 270?

What is the strategy for 1) making sure you have all the solutions? 2) finding those more illusive solutions?

 

[Millennial]

Thanks for your responses. So the other solutions within the given range are 90 and 270?

What is the strategy for 1) making sure you have all the solutions? 2) finding those more illusive solutions?

Yes, those are the solutions.

From your questions, I understand that you didn't get what we are trying to tell you. Consider this rather simple example: [itex]x^2 + x = 0[/itex].

Your erroneous way of solving this equation would be: Divide both sides by x so that you are left with [itex]x + 1 = 0[/itex], and hence [itex]x = -1[/itex]. That is one of the roots, but you can't obtain the second one [itex]x=0[/itex]. The reason for this is you assume x is not 0 when dividing by it, so if you want to do this the division way, the correct solution is this:

"Assume x is not 0. Then, we can divide by x and obtain [itex]x+1=0[/itex], which means we have one root -1. Now, assume x is 0. The equation is satisfied, and hence 0 is the second root."

There is also another way that might be less prone to confusion, suggested by LCKurtz. Factoring the example equation I provided, you get [itex]x(x+1) = 0[/itex]. If the product of two things equals 0, one of them must be 0. Equating x and x+1 to 0, you get the two roots 0 and -1. Note that there is no division in this solution. If the above logic about dividing by zero is confusing for you, the factoring way of doing it might be easier for you, as you don't have to worry about any divisions by zero.

 

[Appleton]

You're right, I didn't get it, but I do now, thanks for clearing that up for me.